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MIT15_053S13_rec07sol

# MIT15_053S13_rec07sol - Optimization Methods in Management...

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Optimization Methods in Management Science MIT 15.053 Recitation 7 TAs: Giacomo Nannicini, Ebrahim Nasrabadi Problem 1 Suppose branch and bound is being applied to a 0–1 integer program in which we are maximizing. By node 11 the first three variables have been fixed. An incumbent has already been found (in another part of the tree) and has value 15. Furthermore, x 4 has been chosen to be the next variable to branch on. This situation is illustrated in the following fragment of the branch and bound tree: 11 12 13 Incumbent: z = 15 z LP ° 16 x 4 = 1 x 4 = 0 x 1 , x 2 , x 3 fixed Figure 1: A fragment of the branch and bound tree for Problem 1. The LP relaxation solved at node 11 (which has x 4 and x 5 as free variables) was: max 18 x 4 + ax 5 + 10 s.t. 8 x 4 + 10 x 5 5 , 0 x 4 1 , 0 x 5 1 , (Hint: For the following questions, the LP relaxation essentially becomes a one variable problem that can be solved by inspection.) Part A.1 Suppose x 4 is set to 1. Does node 12 get pruned (fathomed)? Justify your answer. Solution. Yes, node 12 gets fathomed. If x4 is set to 1, then the LP relaxation becomes:

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max ax 5 + 28 s.t. 10 x 5 ≤ − 3 , 0 x 5 1 , The relaxation is infeasible because of the first constraint, so node 12 gets fathomed. Part B.1 Suppose x 4 is set to 0 and parameter a = 12. Does node 13 get pruned (fathomed)? Justify your answer. Solution. No, node 13 does not get fathomed. If x 4 = 0 and a = 12, then the LP relaxation becomes: max 12 x 5 + 10 s.t. 10 x 5 5 , 0 x 5 1 , The optimal solution to this relaxation is x 5 = 1 / 2, with an objective value of 16. Because the solution is
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